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SQUARING and NOT SQUARING ONE OR MORE PLANES a Square SQUARING AND NOT SQUARING ONE OR MORE PLANES FREDERICK V. HENLE AND JAMES M. HENLE Abstract. A set of natural numbers tiles the plane if a square-tiling of the plane exists using exactly one square of sidelength n for every n in the set. In [9] it is shown that N, the set of all natural numbers, tiles the plane. We answer here a number of questions from that paper. We show that there is a simple tiling of the plane (no non- trivial subset of squares forms a rectangle). We show that neither the odd numbers nor the prime numbers tile the plane. We show that N can tile many, even infinitely many planes. A square-tiling is perfect if no two squares used are the same size. In 1903 Max Dehn [2] asked: Is there a perfect square-tiling of a square? In 1925 Zbigniew Moro´n found perfect square-tilings of several rectangles [10]. Dehn’s question was ultimately answered affirmatively in 1938 by Roland Sprague [12]. The problem and its solution were the subject of a memorable paper, “Squaring the Square” by Tutte [13], reprinted in Martin Gardner’s column in Scientific American (see [6]). Papers have continued to appear on the subject ever since (see for example, [5], [3], [4]). In 1975 Solomon Golomb [7] asked if there was a perfect square-tiling of the infinite plane with every side-length represented. In 1997, Karl Scherer [11] found an imperfect square-tiling of the plane—squares of all integral sides are used, but each size is used multiple times. The number of squares of side n used, s(n), is finite but the function s is not bounded. Golomb’s question was ultimately answered affirmatively in “Squaring the Plane.” (2008, [9]). The solution opened a host of questions, for example, Which sets tile the plane? Is there a three-colorable tiling? Can the half-plane be tiled? There are connections between squaring planes and squaring squares (see for example the proof of 5.1). There are also curious disconnects. There is a clever proof that a cube cannot be cubed ([13]). But the technique has not yet shown us that space cannot be cubed. In section 1, we find a large class of sets, including the set of odd numbers, that do not tile the plane. In section 2, we show that N can tile many planes at once. In section 3, we show that the prime numbers do not tile the plane. In trying to square squares, Tutte and his fellow researchers especially prized “simple” tilings, tilings in which no nontrivial subset of the squares forms a rectangle. One question in [9] asked if there is a simple N-tiling of the plane. In section 4, we construct a simple, perfect, square-tiling of the plane. In section 5, we report on the questions posed in [9], give a Z-tiling of the half-plane, and pose a number of new questions. Since the results in this paper were obtained, additional research has been done, [1], which deepens the mystery of tiling sets. It is now known, for example, that a set with Date: November 19, 2014. 2010 Mathematics Subject Classification. 05B45, 52C20. Key words and phrases. tiling, squares. 1 2 FREDERICK V. HENLE AND JAMES M. HENLE one or three odd numbers may tile the plane but a set with two odd numbers can’t. It also known that a set growing faster than the Fibonacci numbers can’t tile the plane. In this paper, “tiling” will mean a perfect square-tiling. If a tiling T uses all and only the squares in X, we will say T is an “X-tiling”. For simplicity, we will denote the square of side n with the boldface letter ‘n’. 1. A class of sets that do not tile the plane A simple examination of [9] shows the following: Theorem 1.1. Suppose for X ⊆ N that a + b ∈ X for all a, b ∈ X, a =6 b. Then X tiles the plane. In [9] we noted that E, the set of even natural numbers, tiles the plane and asked if O, the set of odd natural numbers, tiled the plane. It turns out that closure under addition is important. E satisfies the hypothesis of . O, on the other hand, is anti-closed. That it fails to tile the plane is a consequence of the following theorem. Theorem 1.2. Suppose for X ⊆ N that for all a, b ∈ X, a =6 b, a + b∈ / X. Then X does not tile the plane. Proof. Suppose there is an X-tiling X . We will derive a contradiction. First note that at every corner of a square in any tiling there will be a third or fourth edge extending from the corner. We will say that a square is a pinwheel if there are lines at the corners for one of these patterns: If one side of the square has lines in the same direction, then we will say it has an integral side (it will have a whole number, n, of neighbors on that side, n> 1). A square may both be a pinwheel and have an integral side, but a square with no integral side must be a pinwheel. Claim 1.3. No square in X can have an integral side. Suppose that the claim is false. Let s be the smallest square with an integral side. Square s can’t have two neighbors on a side because of the assumption on X. Thus, s must have an integral side with at least three neighbors. Along that side, the smallest neighbor a must be at one end, since otherwise, SQUARINGANDNOTSQUARINGONEORMOREPLANES 3 a s a would have an integral number of neighbors along its top edge, contradicting the choice of s as the smallest such square. But now consider a’s neighbor b and suppose that b’s other neighbor is larger. a b s At the corner of b above a there must either be an edge extending up or an edge extending to the left. a b s In either case, b is forced to have an integral side, a contradiction. Finally, if b’s neighbor is smaller, b a s Then the corners of b still present a problem. b a s No matter which way they go b will have an integral side. This proves 1.3. Now consider the smallest square c in X . Square c must be a pinwheel. Let d be the smallest of the surrounding squares. 4 FREDERICK V. HENLE AND JAMES M. HENLE d c The upper left corner of d has an edge going up or to the left. If it goes up, d c we have a problem because the upper right vertex of d will have a third edge coming from it and whichever direction that edge takes, we will have the picture forbidden by the claim. On the other hand, suppose the edge goes to the left. E d c There can’t be a single square in the region marked E, since we would then have two members of X adding to a third member of X. G d f c But square f must be larger than d (d was chosen as the smallest around c), so every square in the region marked G must be smaller than c (chosen as smallest). This is a contradiction and the proof is complete. Corollary 1.4. The set of odd numbers does not tile the plane. 2. Covering several planes The fact that the set of even numbers tiles the plane but its complement does not raises the question: is there a set X such that both X and Xc tile the plane? In fact, there is. Indeed, we can partition N into any finite number of sets, even infinitely many sets, all of which tile the plane. We’ll start with just two. SQUARINGANDNOTSQUARINGONEORMOREPLANES 5 Theorem 2.1. There is a set of natural numbers X such that both X and Xc tile the plane. We need to review here the proof that N tiles the plane. The key is the following lemma. Lemma 2.2. Given any ell-shaped region formed of squares, it is possible to surround the ell with larger squares in such a way as to form a solid rectangle, with the added squares all larger than those in the ell. With this lemma, the proof of the theorem proceeds as follows: (1) Place two squares in the plane, forming an ell. (2) Add squares to the ell to create a rectangle. (3) Add to the rectangle the smallest square not yet used, forming an ell. Steps 2 and 3 are then repeated ad infinitum. Now the proof of 2.1. Proof. In effect, we will show that N can tile two planes, P0 and P1. In stages, we will c c build X, X , tiling P0 with X and P1 with X . The stages will address the planes P0 and P1 alternately. At the end of each stage there will be a finite number of numbers each in X and Xc. The squares with side-lengths in X will be arranged in a rectangle c in P0 and the squares with side-lengths in X will be arranged in a rectangle in P1. We begin by placing two distinct squares, one in each plane. c Suppose we have last addressed X and P0. Here is how we address X and P1. 1. We choose n larger than all the members yet chosen of X and Xc.
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